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3 years ago

Select me correct answer.

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

6 0

The first one. The second looks like (p-6)+2. The third looks like (2-p)6. And the fourth looks like 6+ (2+p)

aalyn [17]3 years ago

5 0

2 - (p + 6)

A. two minus the sum(addition) of p and six(so p and 6 are added together)

2 - (p + 6)

B. two added to the difference(subtraction) of p and six(p an 6 are subtracted together)

2 + (p - 6)

C. the difference of two and p times(multiplication) six

6(2 - p)

D. the sum of two and p added to six

(2 + p) + 6

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0.06 multiplied by 200

horrorfan [7]

Your answer would be 12

5 0

3 years ago

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Suppose y varies directly as x, and y =15 when x = 3. Find the constant of variation (k).

vovangra [49]

Answer:

k = 5

Step-by-step explanation:

y = kx

15 = k(3)

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k = 5

8 0

3 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$